Is the modulus of curvature the speed at which tangents change direction? I'm having trouble understanding an argument: the modulus of curvature it is the speed at which tangents change direction.
Let $\alpha:I \rightarrow \mathbb{R}^2$ a plane curve arc length parametrized and $k(s)$ the curvature of $\alpha$ is $s$. Consider the tangent vectors $\alpha'(s_0)$ and $\alpha'(s_0+h)$, where $s_0 \in I$ is fixed and $s_0 + h \in I$. Denote by $\phi(h)$ the angle between these two vectors, ie,
\begin{equation}
\cos\phi(h) = \langle  \alpha'(s_0) ,\alpha'(s_0+h) \rangle.
\end{equation}
(because $|\alpha'(s_0)| = |\alpha'(s_0+ h)| = 1$).
The limit $\displaystyle \lim_{h \rightarrow 0} \frac{\phi(h)}{h}$ is the speed at which tangents change direction. We have
\begin{equation}
 |\alpha'(s_0+h) - \alpha'(s_0)| = 2 \sin\frac{\phi(h)}{2}
\end{equation}
for all $h$ and so
\begin{equation}
|k(s_0)| = |\alpha''(s_0)| = \lim_{h \rightarrow 0} \frac{\phi(h)}{h}.
\end{equation}
Comments:
I'm not able to verify the last two equations:
$$|\alpha'(s_0+h) - \alpha'(s_0)|^2 = |\alpha'(s_0+h)|^2 - 2 \langle \alpha'(s_0+h) , \alpha'(s_0) \rangle + |\alpha'(s_0)|^2 = 2 - 2\cos{\phi(h)}.$$
I don't know which trigonometric identity is being used.
The last equation is also not being able to verify.
Thank you for your help.
 A: By definition,
$$|\alpha'(s_0+h) - \alpha'(s_0)|^2 = \langle \alpha'(s_0+h) - \alpha'(s_0),\alpha'(s_0+h) - \alpha'(s_0)\rangle .$$
Now apply linearity and symmetry of scalar product (we have real vectors).
In other words: expand.
The vectors are unit, this gives the single $2$ in the final result, the cosine follows from your starting notation.
A: Here is an easier, alternative solution. Without loss of generality, let's assume $\alpha'(s_0)=(1,0)$. Then, by definition, $\alpha'(s_0+h)=(\cos\phi(h),\sin\phi(h))$. Differentiating, we find by the chain rule that
$$|k(s_0+h)| = \|\alpha''(s_0+h)\| = |\phi'(h)|.$$
(By the way, you're missing an absolute value in your final equation for $|k'(s_0)|$. If the curve is concave at $s_0$, the derivative $\phi'(0)$ will be negative.)
A: Yes. As an example let $\alpha(s_o)= \dfrac{s^3}{6 a^2}$ be position vector and let the curve with unit speed slope be
$$\alpha'(s_o)=\phi=\tan^{-1}\dfrac{dy}{dx}=\dfrac{s^2}{2a^2}$$
(for a Clothoid spiral) expressed in terms of arc distance $s$, and  evaluated at this point.
Differentiate with respect to $s$
$$\alpha''(s_o)= \dfrac{\dfrac{d^2y}{dx^2}}{(1+(\dfrac{dy}{dx})^2)^\frac32}=\kappa=\dfrac{d\phi}{ds}=\dfrac{s}{a^2}$$
which is its intrinsic/natural equation.
A: The first equation uses cosine double-angle formula:
$$
\cos\phi=1-2\sin^2{\phi\over2}.
$$
The second equation features a standard trick to compute the limit:
$$
|\alpha''(s_0)|
=\lim_{h\to0}{|\alpha'(s_0+h) - \alpha'(s_0)|\over h} 
= \lim_{h\to0}\left({2\over h} \sin\frac{\phi(h)}{2}\right)
\\
= \lim_{h\to0}\left({\phi(h)\over h}\cdot {\sin(\phi(h)/2)\over\phi(h)/2}\right)
= \lim_{h\to0}\left({\phi(h)\over h}\right),
$$
because $\displaystyle{\sin\theta\over\theta}\to1$ for $\theta\to0$.
